# Bernoulli's inequality

An illustration of Bernoulli's inequality, with the graphs of ${\displaystyle y=(1+x)^{r}}$ and ${\displaystyle y=1+rx}$ shown in red and blue respectively. Here, ${\displaystyle r=3.}$

In real analysis, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x.

The inequality states that

${\displaystyle (1+x)^{r}\geq 1+rx}$

for every integer r ≥ 0 and every real number x ≥ −2. If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads

${\displaystyle (1+x)^{r}>1+rx}$

for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0.

There is also a generalized version that says for every real number r ≥ 1 and real number x ≥ -1,

${\displaystyle (1+x)^{r}\geq 1+rx,}$

while for 0 ≤ r ≤ 1 and real number x ≥ -1,

${\displaystyle (1+x)^{r}\leq 1+rx.}$

Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction, as shown below.

## History

Jacob Bernoulli first published the inequality in his treatise “Positiones Arithmeticae de Seriebus Infinitis” (Basel, 1689), where he used the inequality often.[1]

According to Joseph E. Hofmann, Über die Exercitatio Geometrica des M. A. Ricci (1963), p. 177, the inequality is actually due to Sluse in his Mesolabum (1668 edition), Chapter IV "De maximis & minimis".[1]

## Proof of the inequality

We proceed with mathematical induction in the following form:

• we prove the inequality for ${\displaystyle r\in \{0,1\}}$,
• from validity for some r we deduce validity for r+2.

For r = 0,

${\displaystyle (1+x)^{0}\geq 1+0x}$

is equivalent to 1 ≥ 1 which is true as required.

Similarly, for r = 1 we have

${\displaystyle (1+x)^{r}=1+x\geq 1+x=1+rx.}$

Now suppose the statement is true for r = k:

${\displaystyle (1+x)^{k}\geq 1+kx.}$

Then it follows that

{\displaystyle {\begin{aligned}(1+x)^{k+2}&=(1+x)^{k}(1+x)^{2}\\&\geq (1+kx)\left(1+2x+x^{2}\right)\qquad \qquad \qquad {\text{ by hypothesis and }}(1+x)^{2}\geq 0\\&=1+2x+x^{2}+kx+2kx^{2}+kx^{3}\\&=1+(k+2)x+kx^{2}(x+2)+x^{2}\\&\geq 1+(k+2)x\end{aligned}}}

since ${\displaystyle x^{2}\geq 0}$ as well as ${\displaystyle x+2\geq 0}$. By the modified induction we conclude the statement is true for every non-negative integer r.

## Generalization

The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then

${\displaystyle (1+x)^{r}\geq 1+rx}$

for r ≤ 0 or r ≥ 1, and

${\displaystyle (1+x)^{r}\leq 1+rx}$

for 0 ≤ r ≤ 1.

This generalization can be proved by comparing derivatives. Again, the strict versions of these inequalities require x ≠ 0 and r ≠ 0, 1.

## Related inequalities

The following inequality estimates the r-th power of 1 + x from the other side. For any real numbers xr  with r > 0, one has

${\displaystyle (1+x)^{r}\leq e^{rx},}$

where e = 2.718.... This may be proved using the inequality (1 + 1/k)k < e.

## Alternative form

An alternative form of Bernoulli's inequality for ${\displaystyle t\geq 1}$ and ${\displaystyle 0\leq x\leq 1}$ is:

${\displaystyle (1-x)^{t}\geq 1-xt.}$

This can be proved (for integer t) by using the formula for geometric series: (using y=1-x)

${\displaystyle t=1+1+\dots +1\geq 1+y+y^{2}+\ldots +y^{t-1}={\frac {1-y^{t}}{1-y}}}$

or equivalently ${\displaystyle xt\geq 1-(1-x)^{t}.}$

## Alternative Proof

Using AM-GM

An elementary proof for ${\displaystyle 0\leq r\leq 1}$ and x ≥ -1 can be given using Weighted AM-GM.

Let ${\displaystyle \lambda _{1},\lambda _{2}}$ be two non-negative real constants. By Weighted AM-GM on ${\displaystyle 1,1+x}$ with weights ${\displaystyle \lambda _{1},\lambda _{2}}$ respectively, we get

${\displaystyle {\dfrac {\lambda _{1}\cdot 1+\lambda _{2}\cdot (1+x)}{\lambda _{1}+\lambda _{2}}}\geq {\sqrt[{\lambda _{1}+\lambda _{2}}]{(1+x)^{\lambda _{2}}}}}$

Note that

${\displaystyle {\dfrac {\lambda _{1}\cdot 1+\lambda _{2}\cdot (1+x)}{\lambda _{1}+\lambda _{2}}}={\dfrac {\lambda _{1}+\lambda _{2}+\lambda _{2}x}{\lambda _{1}+\lambda _{2}}}=1+{\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}x}$

and

${\displaystyle {\sqrt[{\lambda _{1}+\lambda _{2}}]{(1+x)^{\lambda _{2}}}}=(1+x)^{\frac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}}$

so our inequality is equivalent to

${\displaystyle 1+{\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}x\geq (1+x)^{\frac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}}$

After substituting ${\displaystyle r={\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}}$ (bearing in mind that this implies ${\displaystyle 0\leq r\leq 1}$) our inequality turns into

${\displaystyle 1+rx\geq (1+x)^{r}}$ which is Bernoulli's inequality.

Using the formula for geometric series

Bernoulli's inequality

${\displaystyle (1+x)^{r}\geq 1+rx}$

(1)

is equivalent to

${\displaystyle (1+x)^{r}-1-rx\geq 0}$

(2)

and by the formula for geometric series (using y=1+x) we get

${\displaystyle (1+x)^{r}-1=y^{r}-1=\left(\sum _{k=0}^{r-1}y^{k}\right)\cdot (y-1)=\left(\sum _{k=0}^{r-1}(1+x)^{k}\right)\cdot x}$

(3)

${\displaystyle (1+x)^{r}-1-rx=\left(\left(\sum _{k=0}^{r-1}(1+x)^{k}\right)-r\right)\cdot x=\left(\sum _{k=0}^{r-1}\left((1+x)^{k}-1\right)\right)\cdot x\geq 0}$

(4)

Now if ${\displaystyle x\geq 0}$ then by monotony of the powers each summand ${\displaystyle (1+x)^{k}-1\geq 0}$, therefore their sum is greater ${\displaystyle 0}$ and hence the product on the LHS of (4).

If ${\displaystyle 0\geq x\geq -2}$ then by the same arguments ${\displaystyle 1\geq (1+x)^{k}}$ and thus all addends ${\displaystyle (1+x)^{k}-1}$ are non-positive and hence their sum. Since the product of two non-positive numbers is non-negative, we get again (4).

Using Binomial theorem

(1) For x > 0, ${\displaystyle (1+x)^{r}=1+rx+{\tbinom {r}{2}}x^{2}+...+{\tbinom {r}{r}}x^{r}}$ Obviously, ${\displaystyle {\tbinom {r}{2}}x^{2}+...+{\tbinom {r}{r}}x^{r}\geq 0}$

Thus, ${\displaystyle (1+x)^{r}\geq 1+rx}$

(2) For x = 0, ${\displaystyle (1+x)^{r}=1+rx}$

(3) For −1 ≤ x < 0, let y = −x, then 0 < y ≤ 1

Replace x with −y, we have ${\displaystyle (1-y)^{r}=1-ry+{\tbinom {r}{2}}y^{2}+...+(-1)^{r}{\tbinom {r}{r}}y^{r}}$

Also, according to the binomial theorem, ${\displaystyle 0=1-r+{\tbinom {r}{2}}+...+(-1)^{r}{\tbinom {r}{r}}}$

then${\displaystyle {\tbinom {r}{2}}+...+(-1)^{r}{\tbinom {r}{r}}=r-1\geq 0}$

Notice that ${\displaystyle y^{2}\geq y^{3}\geq ...\geq y^{r}}$

Therefore, we can see that each binomial term ${\displaystyle {\tbinom {r}{n}}}$ is multiplied by a factor ${\displaystyle y^{n}}$ , and that will make each term smaller than the term before.

For that reason, ${\displaystyle {\tbinom {r}{2}}y^{2}+...+(-1)^{r}{\tbinom {r}{r}}y^{r}\geq 0}$

Hence, ${\displaystyle (1-y)^{r}=1-ry+{\tbinom {r}{2}}y^{2}+...+(-1)^{r}{\tbinom {r}{r}}y^{r}\geq 1-ry}$

Replace y with −x back, we get ${\displaystyle (1+x)^{r}\geq 1+rx}$

Notice that by using binomial theorem, we can only prove the cases when r is a positive integer or zero.

## References

• Carothers, N.L. (2000). Real analysis. Cambridge: Cambridge University Press. p. 9. ISBN 978-0-521-49756-5.
• Bullen, P. S. (2003). Handbook of means and their inequalities. Dordercht [u.a.]: Kluwer Academic Publ. p. 4. ISBN 978-1-4020-1522-9.
• Zaidman, S. (1997). Advanced calculus : an introduction to mathematical analysis. River Edge, NJ: World Scientific. p. 32. ISBN 978-981-02-2704-3.